CAT 2019 Question Paper Slot 1

Assume the average of 21 students other than Ramesh = a

Sum of the scores of 21 students other than Ramesh = 21a

Hence the average of 22 students = a+1

Sum of the scores of all 22 students = 22(a+1)

The score of Ramesh = Sum of scores of all 22 students - Sum of the scores of 21 students other than Ramesh = 22(a+1)-21a=a+22  = 82.5   (Given)

=> a = 60.5

Hence, sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353

Now the sum of the scores of students other than Gautam = 21*62 = 1302

Hence the score of Gautam = 1353-1302=51

We have, $$(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$$

Converting both sides in powers of 2 and 3, we get

$$2^{\ \frac{19\ }{2}}3^42^43^{2m}2^{3n}$$ = $$3^n2^{4m}2^{\frac{\ 6}{4}}$$

Comparing the power of 2 we get, $$\ \frac{\ 19}{2}+4+3n\ =4m+\frac{\ 6}{4}\ $$

=> 4m=3n+12 .....(1)

Comparing the power of 3 we get, $$4+2m=n$$

Substituting the value of n in (1), we get

4m=3(4+2m)+12

=> m=-12

Consider the work done by a man in a day = a and that by a machine = b

Since, three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job, hence the efficiency will be double.

=> 3a+8b = 2(3b+8a)

=> 13a=2b

Hence work done by 13 men in a day = work done by 2 machines in a day.

=> If two machines can finish the job in 13 days, then same work will be done 13 men in 13 days. 

Hence the required number of men = 13

The given figure can be divided into 9 regions or equilateral triangles of equal areas as shown below,

Now the hexagon consists of 6 regions and the triangle consists of 9 regions.

Hence the ratio of areas = 6/9 =2:3

We have, $$\log_{5}{(x + y)} + \log_{5}{(x - y)} = 3$$

=> $$x^2-y^2=125$$......(1)

$$\log_{2}{y} - \log_{2}{x} = 1 - \log_{2}{3}$$

=>$$\ \frac{\ y}{x}$$ = $$\ \frac{\ 2}{3}$$

=> 2x=3y   => x=$$\ \frac{\ 3y}{2}$$

On substituting the value of x in 1, we get

$$\ \frac{\ 5x^2}{4}$$=125

=>y=10, x=15

Hence xy=150

Sum of the area of region I and II is the required area.

Now, required area =  $$\ 4\times\frac{\ 1}{2}\times\ 1\times\ 1$$ = 2

The number of paths from (1, 1) to (8, 10) via (4, 6) = The number of paths from (1,1) to (4,6) * The number of paths from (4,6) to (8,10)

To calculate the number of paths from (1,1) to (4,6), 4-1 =3 steps in x-directions and 6-1=5 steps in y direction

Hence the number of paths from (1,1) to (4,6) = $$^{(3+5)}C_3$$ = 56

To calculate the number of paths from (4,6) to (8,10), 8-4 =4 steps in x-directions and 10-6=4 steps in y direction

Hence the number of paths from (4,6) to (8,10) = $$^{(4+4)}C_4$$ = 70

The number of paths from (1, 1) to (8, 10) via (4, 6) = 56*70=3920

Assuming the dimensions of the brick are a, b and c and the diagonals are 3, 2 $$\surd3$$ and $$\surd{15}$$

Hence, $$a^{2\ }+\ b^2$$ = $$3^2$$   ......(1)

$$b^{2\ }+\ c^2$$ = $$(2\sqrt{3})^2$$ ......(2)

$$c^{2\ }+\ a^2$$ = $$(\sqrt{15})^2$$ ......(3)

Adding the three equations, 2($$a^2+b^2+c^2$$) = 9+12+15=36

=>$$a^2+b^2+c^2$$ = 18......(4)

Subtracting (1) from (4), we get $$c^2$$ = 9    =>c=3

Subtracting (2) from (4), we get $$a^2$$ = 6    =>a=$$\sqrt{6}$$

Subtracting (3) from (4), we get $$b^2$$ = 3    =>b=$$\sqrt{3}$$

The ratio of the length of the shortest edge of the brick to that of its longest edge is = $$\ \frac{\ \sqrt{3}}{3}$$ = $$1 : \sqrt{3}$$

For x <0, -x($$6x^2+1$$) = $$5x^2$$

=> ($$6x^2+1$$) = -5x

=> ($$6x^2 + 5x+ 1$$) = 0 

=>($$6x^2 + 3x+2x+ 1$$) = 0

=> (3x+1)(2x+1)=0    =>x=$$\ -\frac{\ 1}{3}$$  or x=$$\ -\frac{\ 1}{2}$$

For x=0, LHS=RHS=0    (Hence, 1 solution)

For x >0, x($$6x^2+1$$) = $$5x^2$$

=> ($$6x^2 - 5x+ 1$$) = 0 

=>(3x-1)(2x-1)=0    =>x=$$\ \frac{\ 1}{3}$$   or   x=$$\ \frac{\ 1}{2}$$

Hence, the total number of solutions = 5

In any right triangle, the circumradius is half of the hypotenuse. Here,L=$$\ \frac{\ 1}{2}\ $$* the length of the hypotenuse = $$\ \frac{\ 1}{2}$$($$\sqrt{\ 15^2+9^2}$$) = $$\ \frac{\ 1}{2}\ $$*$$\sqrt{\ 306}$$ = $$\ \frac{\ 1}{\ 2}\times\ $$17.49 = 8.74

Hence, the integer close to L = 9